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If $\alpha, \beta, \gamma$, are the roots of the equation $x^3+4 x+1=0$, then $(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$ is equal to
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4
Since, $\alpha, \beta, \gamma$ are the roots of the equation $x^3+4 x+1=0$
$\therefore \quad \alpha+\beta+\gamma=0$
$\alpha \beta+\beta \gamma+\gamma \alpha=4$
and $\quad \alpha \beta \gamma=-1$
$\therefore(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$
$=\frac{1}{\alpha+\beta}+\frac{1}{\beta+\gamma}+\frac{1}{\gamma+\alpha}$
$=\frac{1}{-\gamma}+\frac{1}{-\alpha}+\frac{1}{-\beta}=-\left[\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}\right]$
$=-\left[\frac{4}{-1}\right]=4$
$\therefore \quad \alpha+\beta+\gamma=0$
$\alpha \beta+\beta \gamma+\gamma \alpha=4$
and $\quad \alpha \beta \gamma=-1$
$\therefore(\alpha+\beta)^{-1}+(\beta+\gamma)^{-1}+(\gamma+\alpha)^{-1}$
$=\frac{1}{\alpha+\beta}+\frac{1}{\beta+\gamma}+\frac{1}{\gamma+\alpha}$
$=\frac{1}{-\gamma}+\frac{1}{-\alpha}+\frac{1}{-\beta}=-\left[\frac{\alpha \beta+\beta \gamma+\gamma \alpha}{\alpha \beta \gamma}\right]$
$=-\left[\frac{4}{-1}\right]=4$
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